System: one point charge Q above a grounded conducting plane
𝑦
𝜕2𝑉 𝜕2𝑉 𝜕2𝑉
Formal approach: 𝛁 𝑉 = 2 + 2 + 2 = 0
𝜕𝑥
𝜕𝑦
𝜕𝑧
2
Conditions for the solution:
1. At all points on the ground conducting
plane, the potential is zero:
𝑄(0, 𝑑, 0)
Grounded plane
conductor
𝑥
𝑧
𝑉 𝑥, 0, 𝑧 = 0
2. At points very close to Q the potential approaches that of the point charge alone:
𝑉→
𝑄
, 𝑎𝑠 𝑅 → 0
4𝜋𝜖0 𝑅
where 𝑅 is the distance from the probe to the source 𝑄
3. At points very far 𝑄(𝑥 → ±∞, 𝑦 → ±∞, 𝑜𝑟 𝑧 → ±∞) the potential approaches zero.
4. The potential function is even with respect to the 𝑥 and 𝑧 coordinates:
𝑉 𝑥, 𝑦, 𝑧 = 𝑉 −𝑥, 𝑦, 𝑧
and
𝑉 𝑥, 𝑦, 𝑧 = 𝑉 𝑥, 𝑦, −𝑧 .
This problem is relatively easy to solve numerically, but an analytical solution is arduous.
The presence of a positive charge 𝑄 at 𝑦 = 𝑑 would induce negative charges on the surface
of the conducting plane, resulting in a surface charge density 𝜌𝑠 . Hence the potential at
points above the conducting plane would be:
𝑉 𝑥, 𝑦, 𝑧 =
𝑄
4𝜋𝜖0
𝑥2
+ (𝑦 −
𝑑)2 +𝑧 2
+
1
4𝜋𝜖0
𝜌𝑠
𝑑𝑠
𝑆 𝑅1
difficult to compute
𝑦
conducting plane
𝑦
𝑄(0, 𝑑, 0)
𝑥
𝑥
 We replace the bounding surface with an appropriate image charge
−𝑄(0, −𝑑, 0)
image charge
 The image charges are located outside the region in which the field is to be determined
 The field in the region of the image charges does not model the original system
𝑦
In the 𝑦 > 0 region is
𝑉 𝑥, 𝑦, 𝑧 =
𝑄(0, 𝑑, 0)
𝑄
1
1
−
4𝜋𝜖0 𝑅+ 𝑅−
𝑥
𝑧
where:
2
2
1
2 2
+𝑧
2
2
1
+𝑧 2 2
𝑅+ = 𝑥 + (𝑦 − 𝑑)
𝑅− = 𝑥 + (𝑦 + 𝑑)
𝑃(𝑥, 𝑦, 𝑧)
𝑅−
𝑄
𝑅+
O
𝑦 = 0 plane
−𝑄
 Models a system with charges and bounding surfaces
through an equivalent system made only with charges
 The original system is correctly modeled only in the region
where there are no images
 Difficult to generalize to geometrically complex systems
 The numerical solution of Poisson’s equation is a convenient
alternative to this method